Higher June 2023 Paper 2 Q12
12 The diagram shows right-angled triangle \(ABD\)
Diagram NOT accurately drawn
\(AB = 14\) cm \(\qquad AD = 8\) cm
\(C\) is the point on \(BD\) such that angle \(BAC = 38^\circ\)
Work out the length of \(CD\)
Give your answer correct to 3 significant figures.
(4)
| Scheme | Marks |
|---|---|
\(\cos BAD = \dfrac{8}{14}\) or \(\sin ABD = \dfrac{8}{14}\) or \(\sin ABD = \dfrac{8\sin 90}{14}\) or \((BD =)\sqrt{14^2 - 8^2}\left(= \sqrt{132} = 2\sqrt{33} = 11.4(89...)\right)\) | M1 |
\(BAD = \cos^{-1}\left(\dfrac{8}{14}\right)\) (= 55.(15…)) or \(\cos^{-1}\left(\dfrac{14^2 + 8^2 - \text{``}{11.489}\text{''}^2}{2 \times 14 \times 8}\right)\) \(BAD = \sin^{-1}\left(\dfrac{\text{``}{11.489...}\text{''}}{14}\right)\) (= 55.(15…)) or \(BAD = \tan^{-1}\left(\dfrac{\text{``}{11.489...}\text{''}}{8}\right)\) (= 55.(15…)) or \(BAD = 180 - 90 - \sin^{-1}\left(\dfrac{8}{14}\right)\) (= 180 – 90 – 34.8... = 55.(15...)) oe or \(CAD = 180 - 38 - \sin^{-1}\left(\dfrac{8}{14}\right) - 90\) (= 180 – 38 – 34.8 – 90 = 17.2) | M1 |
\(\tan(\text{``}{55.15...}\text{''} - 38) = \dfrac{CD}{8}\) oe eg \(\tan 17.2 = \dfrac{CD}{8}\) \(\dfrac{CD}{\sin(55.1... - 38)} = \dfrac{8}{\sin(90 - (55.1... - 38))}\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 2.47 | A1 |
| (4) | |
| (4 marks) |
Notes
M1: A correct equation with \(CD\) being the only unknown value
A1: 2.44 – 2.48